You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

Example 1:

Input: nums = [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: nums = [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
             Total amount you can rob = 2 + 9 + 1 = 12.

Constraints:

  • 0 <= nums.length <= 100
  • 0 <= nums[i] <= 400

Solution

Time complexity : O(n)
Space complexity : O(1)

class Solution {
public:
    int rob(vector<int>& nums) {
        int n = nums.size();
        if (0 == n) return 0;
        
        int pre1 = 0, pre2 = 0;
        for (int i=0; i<n; ++i) {
            int so_far = max(pre2 + nums[i], pre1);
            pre2 = pre1;
            pre1 = so_far;
        }
        return pre1;
    }
};

從頭開始，每經過一間房子，考慮：

  1. 要搶這間，則目前最大收益為「直到前兩間的最大收益+這間收益」。
  2. 不搶這間，則目前最大收益為「直到前一間的最大收益」。

這兩種考慮中取最大者，一路考慮到最後。